The scalar field $f(x, y) = 3x^3y - x^2y^3 + 4$ has a critical point at $(0, 1)$. How does the second partial derivative test classify this critical point? Choose 1 answer: Choose 1 answer: (Choice A) A Local maximum (Choice B) B Local minimum (Choice C) C Saddle point (Choice D) D The test is inconclusive
The second partial derivative test uses the quantity below, evaluated at the critical point we wish to classify. $H = f_{xx}f_{yy} - f_{xy}f_{yx}$ $H < 0$ implies a saddle point. $H > 0$ and $f_{xx} > 0$ implies a local minimum. $H > 0$ and $f_{xx} < 0$ implies a local minimum. $H = 0$ means the test is inconclusive. Let's calculate $H$. First we need all the regular partial derivatives. $\begin{aligned} f_x &= 9x^2y - 2xy^3 \\ \\ f_y &= 3x^3 - 3x^2y^2 \end{aligned}$ Now we can find all the second order partial derivatives. $\begin{aligned} f_{xx} &= 18xy - 2y^3 = -2 \\ \\ f_{yx} &= 9x^2 - 6xy^2 = 0 \\ \\ f_{xy} &= 9x^2 - 6xy^2 = 0 \\ \\ f_{yy} &= -6x^2y = 0 \end{aligned}$ Therefore, $H = (-2)(0) - (0)(0) = 0$. Because $H$ is zero, the test is inconclusive.